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                <h1 class="title-article">181118 逆向-HXB2018（Reverse）</h1>
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                                                                                                                                            <a class="follow-nickName" href="https://me.csdn.net/whklhhhh" target="_blank" rel="noopener">奈沙夜影</a>
                    <span class="time">2018-11-19 00:27:33</span>
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                                <div class="up-time"><span>最后发布:2018-11-19 00:27:33</span><span>首发:2018-11-19 00:27:33</span></div>
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                                            <p>这次的re难度不是太大……但是re2和re3都有点偏门，不太硬核233 但也挺有意思的</p>
<h1><a id="Replace_2"></a>Replace</h1>
<p><code>upx -d</code>脱壳，然后是一个比普通签到略复杂一点的签到题，没什么好说的<br>
要求<code>table[input[i]] == atoi(data[2*i]+data[2*i+1])^0x19</code></p>
<pre><code>table = [0x63, 0x7C, 0x77, 0x7B, 0xF2, 0x6B, 0x6F, 0xC5, 0x30, 0x01,
  0x67, 0x2B, 0xFE, 0xD7, 0xAB, 0x76, 0xCA, 0x82, 0xC9, 0x7D,
  0xFA, 0x59, 0x47, 0xF0, 0xAD, 0xD4, 0xA2, 0xAF, 0x9C, 0xA4,
  0x72, 0xC0, 0xB7, 0xFD, 0x93, 0x26, 0x36, 0x3F, 0xF7, 0xCC,
  0x34, 0xA5, 0xE5, 0xF1, 0x71, 0xD8, 0x31, 0x15, 0x04, 0xC7,
  0x23, 0xC3, 0x18, 0x96, 0x05, 0x9A, 0x07, 0x12, 0x80, 0xE2,
  0xEB, 0x27, 0xB2, 0x75, 0x09, 0x83, 0x2C, 0x1A, 0x1B, 0x6E,
  0x5A, 0xA0, 0x52, 0x3B, 0xD6, 0xB3, 0x29, 0xE3, 0x2F, 0x84,
  0x53, 0xD1, 0x00, 0xED, 0x20, 0xFC, 0xB1, 0x5B, 0x6A, 0xCB,
  0xBE, 0x39, 0x4A, 0x4C, 0x58, 0xCF, 0xD0, 0xEF, 0xAA, 0xFB,
  0x43, 0x4D, 0x33, 0x85, 0x45, 0xF9, 0x02, 0x7F, 0x50, 0x3C,
  0x9F, 0xA8, 0x51, 0xA3, 0x40, 0x8F, 0x92, 0x9D, 0x38, 0xF5,
  0xBC, 0xB6, 0xDA, 0x21, 0x10, 0xFF, 0xF3, 0xD2, 0xCD, 0x0C,
  0x13, 0xEC, 0x5F, 0x97, 0x44, 0x17, 0xC4, 0xA7, 0x7E, 0x3D,
  0x64, 0x5D, 0x19, 0x73, 0x60, 0x81, 0x4F, 0xDC, 0x22, 0x2A,
  0x90, 0x88, 0x46, 0xEE, 0xB8, 0x14, 0xDE, 0x5E, 0x0B, 0xDB,
  0xE0, 0x32, 0x3A, 0x0A, 0x49, 0x06, 0x24, 0x5C, 0xC2, 0xD3,
  0xAC, 0x62, 0x91, 0x95, 0xE4, 0x79, 0xE7, 0xC8, 0x37, 0x6D,
  0x8D, 0xD5, 0x4E, 0xA9, 0x6C, 0x56, 0xF4, 0xEA, 0x65, 0x7A,
  0xAE, 0x08, 0xBA, 0x78, 0x25, 0x2E, 0x1C, 0xA6, 0xB4, 0xC6,
  0xE8, 0xDD, 0x74, 0x1F, 0x4B, 0xBD, 0x8B, 0x8A, 0x70, 0x3E,
  0xB5, 0x66, 0x48, 0x03, 0xF6, 0x0E, 0x61, 0x35, 0x57, 0xB9,
  0x86, 0xC1, 0x1D, 0x9E, 0xE1, 0xF8, 0x98, 0x11, 0x69, 0xD9,
  0x8E, 0x94, 0x9B, 0x1E, 0x87, 0xE9, 0xCE, 0x55, 0x28, 0xDF,
  0x8C, 0xA1, 0x89, 0x0D, 0xBF, 0xE6, 0x42, 0x68, 0x41, 0x99,
  0x2D, 0x0F, 0xB0, 0x54, 0xBB, 0x16]
s = bytes.fromhex("2a49f69c38395cde96d6de96d6f4e025484954d6195448def6e2dad67786e21d5adae6")
for i in range(len(s)):
    v = table.index(s[i]^0x19)
    print(chr(v), end='')
</code></pre>
<h1><a id="HighwayHash64_38"></a>HighwayHash64</h1>
<p>从题目和描述的Hash，以及输入提示的</p>
<blockquote>
<p>Note:hxb2018{digits}</p>
</blockquote>
<p>就可以猜到，这估计是个爆破Hash的题目</p>
<p>看了一下hash函数中初始化结构体的部分跟md5不同，查了也没有信息，所以可能是自定义的哈希算法<br>
刚开始尝试了一下扒代码到编译器中复现，然而有很多ROL的宏定义，比较麻烦，所以直接调用该函数是比较方便的</p>
<p>调用函数有两种方法，一种是写一个dll注入到exe中进行调用，另一种则是将该exe直接改成dll，另外写一个exe来调用<br>
前者日后再尝试吧，相对而言感觉要复杂一些<br>
后者只需要将exe的PE头中的标志位修改，再通过RVA(Relative Virtual Address)获取函数地址即可</p>
<p>具体方法为，首先通过十六进制编辑器修改PE头<br>
这里使用010Editor一类的工具会比较方便</p>
<ul>
<li>NtHeader
<ul>
<li>Characteristics
<ul>
<li>IMAGE_FILE_DLL标志位</li>
</ul>
</li>
</ul>
</li>
</ul>
<p>将该位改为1即可通过<code>LoadLibrary</code>调用</p>
<pre><code>typedef __int64(__fastcall *f)(__int64 buff, unsigned __int64 len);

int main()
{
	HINSTANCE hdll;

	hdll = LoadLibrary(TEXT("F:\\ctf\\hxb\\2018\\reverse.dll"));
	if (hdll == NULL)
	{
		printf("Load dll Error: %d\n", GetLastError());
		return 0;
	}
	printf("Dll base is %llx\n", hdll);
	func = ((f)((char*)hdll + 0x17A0));
}
</code></pre>
<p>注意编译的时候由于dll是x64的，因此exe理应也是用x64的<br>
以及这里的函数声明需要使用<code>__fastcall</code>的调用约定，因为从汇编可以看出来</p>
<pre><code>mov     edx, 4
mov     [rsp+158h+var_138], eax
lea     rcx, [rsp+158h+var_138]
call    hash
</code></pre>
<p>传参使用的rcx和rdx，如果用其他调用约定的话通常会用栈传参<br>
IDA其实是已经识别出来的<br>
返回值则需要自己根据内容看出来，向rax放了一个int64的值</p>
<pre><code>mov     rax, qword ptr [rsp+0C8h+md5_struct]
add     rax, qword ptr [rsp+0C8h+md5_struct+20h]
add     rax, qword ptr [rsp+0C8h+md5_struct+40h]
add     rax, qword ptr [rsp+0C8h+md5_struct+60h]
</code></pre>
<p>这里IDA是识别错误的</p>
<p>接下来就可以直接使用该函数来爆破了</p>
<pre><code> do
    ++len;
  while ( Dst[len] );
  v7 = len;
  if ( hash((__int64)&amp;v7, 4ui64) != (char *)0xD31580A28DD8E6C4i64 )
</code></pre>
<p>第一次hash使用的是len的地址，也就是把长度视作一个4字节的char数组来进行hash<br>
因此我们首先要算出flag的长度<br>
爆破的时候也提供一个int的空间即可</p>
<pre><code>void len()
{
	int i;
	unsigned long long  result;
	for (i = 0;i&lt;50; i++)
	{
		result = func((long long )&amp;i, 4);
		if (result == 0xD31580A28DD8E6C4)
		{
			printf("Len is %d\n", i-9);
			return ;
		}
	}
	printf("Not found the len\n");
	return;
}
</code></pre>
<p>很快可以得出i=19，然后掐去前后的格式字符共9个，即可知道中间的内容是十个十进制数了</p>
<p>接下来可以通过sprintf快速制作10个字节的十进制数，然后穷举</p>
<pre><code>void hash()
{
	unsigned long long i;
	unsigned long long  result;
	char buff[20];
	for (i = 0; i &lt; 10000000000; i++)
	{
		sprintf_s(buff, "%0.10llu", i);
		if (i % 100000 == 0)
		{
			printf("%0.10llu\n", i);
		}
		result = func((long long)buff, 10);
		if (result == 0x7CDCCF71350B7DB8)
		{
			//5203614978
			printf("flag is %lld\n", i);
			return;
		}
	}
}
</code></pre>
<p><img src="https://i.imgur.com/pdC2p5P.png" alt=""><br>
赛后交流了一下，flag的hash是不一样，所以复现的时候需要自己改一下</p>
<h1><a id="More_efficient_than_JS_154"></a>More efficient than JS</h1>
<p>题目文件下载下来就能看到一个wasm，再结合题目，很显然又是WebAssembly逆向……越来越多的出题人开始搞这东西了orz<br>
目前Chrome和Firefox都没有针对它出好用的调试器，只能用js的调试器凑活看，所以我的经验就是直接动调，用wabt组件反编译出的c和wat来辅助分析</p>
<p>运行了一下直接在fetch的地方报错，让队里师傅给搭了个http环境才能跑起来</p>
<p>以往的wasm题目都是在html中调用函数，这次找了一圈也没有看到<br>
跑起来以后什么都不显示直接弹窗，估计是js中的代码，于是根据提示内容"Input:"找到了这里<br>
<img src="https://i.imgur.com/wkrsFl3.png" alt=""><br>
在这下断，然后刷新果然断到了，但是接着单步跟下去就会进到一个死循环里<br>
<img src="https://i.imgur.com/9dGo3Pi.png" alt=""><br>
这个循环执行完以后又会回到弹窗里，于是有点懵逼</p>
<p>（动态调试和静态分析的相关技巧可以在我前两天的<a href="https://blog.csdn.net/whklhhhh/article/details/84146188">博客</a>中找到）<br>
后来在wat里发现了一个函数的名字叫做<code>_main</code><br>
果断下断，发现刷新页面以后会先执行wasm中的<code>_main</code>函数，然后到<code>f98</code>的调用时开始弹窗，点击取消以后会继续执行<br>
再往后两个调用，到<code>f42</code>的时候注意它的参数执行完后会取出值，返回值则是len</p>
<p>然后在f22的5个参数中就可以发现各种有趣的东西了，注释如下<br>
<img src="https://i.imgur.com/48Wm0sb.png" alt=""><br>
其中f22是核心的加密函数，在里面不断地单步跟，有选择地跳过<br>
（其实最关键的就是几个循环中的i32_load），看它们从哪里取值以及取出来的是什么值即可</p>
<p>个人认为wasm的关键在于跟随数据而不是代码（因为代码太恶心了orz）<br>
中间可以看到根据key去往后table中取值，但是最后与输入有关的只是异或，因此可以输入一串0，从而得到异或的值<br>
然后从f23中的一个循环中使用的地址得到结果数组，最后异或求解即可</p>
<pre><code>input = [[137, 221, 46, 119, 76, 156, 92, 92, 137, 215, 225, 85, 132, 233, 53, 206, 231, 78, 160, 89, 133, 178, 65, 60, 63, 29, 11, 164, 233, 71, 5, 192, 227, 190, 31, 178, 177, 218, 213, 38, 217, 39, 137, 164, 117, 224]]
output = [223, 129, 127, 32, 7, 196, 13, 28, 201, 158, 142, 23, 215, 237, 120]
for i in range(len(output)):
    print(chr(input[i]^ord('0')^output[i]),end='')
</code></pre>
<blockquote>
<p>flag{happy_rc4}</p>
</blockquote>
<p>从这个flag来看算法应该是rc4，也比较负责动调中感觉到的，根据key变换table然后取table的值和明文异或<br>
因为这个算法的特性所以也可以理解为和密钥流异或23333</p>

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